bash shell display first part of file using BSD head

Question

Here's the file:

line1
line2
⋮
line10
Total
Update time

I want to have the first part of file except for the last 2 lines.

The output should look like this:

line1
line2
⋮
line10

How can i achieve this using BSD head? or other oneliner?

Thanks!

Take a look at [one-liner: print all lines except the last 3?](http://stackoverflow.com/questions/18923097/one-liner-print-all-lines-except-the-last-3) — Birei, Sep 21 '13 at 19:23

user1502952 · Accepted Answer · 2013-09-21T19:52:58.467

2

using awk, this will print all the lines except last two lines.

  awk '{a[j++]=$0}END{for(i=0;i<j-2;i++){print a[i]}}' filename

edited Sep 21 '13 at 19:52

answered Sep 21 '13 at 19:43

user1502952

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need to change i=1 to i=0. (: – sbs Sep 21 '13 at 19:48

score 2 · Answer 2 · answered Sep 21 '13 at 19:55

2

Another awk option:

cat <<file>> | awk -v len=$(wc -l <<file>> | awk '{print $1}') 'NR < len - 2 { print $0 }'

answered Sep 21 '13 at 19:55

user2141650

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score 2 · Answer 3 · answered Sep 21 '13 at 21:23

2

reverse the file, delete the first 2 lines, then reverse back:

tac file | sed 1,2d | tac

answered Sep 21 '13 at 21:23

glenn jackman

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konsolebox · Answer 4 · 2013-09-21T22:43:59.490

The concept of this doesn't use much buffers and is useful even for extremely large files or endless streams in pipes. It should be better than any other solution provided so far. You also don't have to wait to get to the end before it starts printing the lines.

awk 'BEGIN{X = 2; getline; a[NR % X] = $0; getline; a[NR % X] = $0}{print a[NR % X]; a[NR % X] = $0}'

NR could also be replaced with another counter that resets itself to 0 if limit of NR is concerned. Also it would depend on the implementation of awk how it cleans itself up and reuse memory on every N lines of input but the concept is there already.

awk 'BEGIN{X = 2; for(i = 0; i < X; ++i)getline a[i]}{i %= X; print a[i]; a[i++] = $0}'

score 1 · Answer 5 · answered Sep 21 '13 at 19:24

Something like

head file.txt -n -2

should work, provided your head version allows the -2 argument. My version of GNU's head works

$ head --version
head (GNU coreutils) 5.97
Copyright (C) 2006 Free Software Foundation, Inc.
This is free software.  You may redistribute copies of it under the terms of
the GNU General Public License .
There is NO WARRANTY, to the extent permitted by law.

Written by David MacKenzie and Jim Meyering.

score 1 · Answer 6 · answered Sep 21 '13 at 20:29

1

This might work for you:

[root@gc-rhel6-x64 playground]# cat file
line1
line2
line3
line4
line5
line6
line7
line8
line9
line10

[root@gc-rhel6-x64 playground]# head -`expr \`wc -l < file\` - 2` file
line1
line2
line3
line4
line5
line6
line7
line8

answered Sep 21 '13 at 20:29

Technext

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use `$()` instead of backticks: `head -$(expr $(wc -l < file) - 2) file` – glenn jackman Sep 21 '13 at 21:21
Thanks @glennjackman! Your way is definitely more readable :) – Technext Sep 22 '13 at 07:37

score 0 · Answer 7 · answered Sep 21 '13 at 19:45

0

This will get the file except last 2 lines

cat file.txt | head -$(expr $(wc -l file.txt | cut -c-8) - 2)

change the bold part for modification.

answered Sep 21 '13 at 19:45

sbs

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bash shell display first part of file using BSD head

7 Answers7